Final answer:
In a cross of true-breeding garden pea plants with green seeds and yellow seeds, the F1 offspring will have 100 percent yellow seeds, as yellow is the dominant allele. The genotype of all F1 offspring will be heterozygous Yy.
Step-by-step explanation:
When performing a cross involving garden pea plants for seed color, yellow seed color is dominant over green seed color. Considering we have true-breeding parents, one with green seeds and one with yellow seeds, all F1 offspring would express the dominant phenotype. Using a Punnett square, we can illustrate that a parent with green seeds (recessive) would have the genotype yy, and a parent with yellow seeds (dominant) would have the genotype YY. During gamete formation, the genotypes segregate, resulting in Y gametes from the yellow-seeded parent and y gametes from the green-seeded parent.
Upon fertilization, each offspring receives one allele from each parent. This means that all F1 offspring have the genotype Yy, resulting in a phenotype of yellow seeds. Therefore, in the F1 generation, you would expect 100 percent yellow seeds. There would be no green seeds or intermediate seeds, as the dominant yellow allele masks the expression of the recessive green allele in the heterozygous condition. This follows Mendel's law of dominance and illustrates a typical monohybrid cross with a 3:1 phenotypic ratio in the F2 generation, but in the F1 generation, all offspring display the dominant phenotype.